Example

Question:

Slowing down of neutrons: In a nuclear reactor, a neutron of high speed (typically \(10^7\,\mathrm{m\,s^{-1}}\)) must be slowed to \(10^3\,\mathrm{m\,s^{-1}}\) so that it can have a high probability of interacting with isotope \(^{235}U\) and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (\(D_2O\)) or graphite, is called a moderator.

Solution:

The initial kinetic energy of the neutron is: \[ K_{1i} = \frac{1}{2}m_1 v_{1i}^2 \] From elastic collision theory, the final kinetic energy is: \[ K_{1f} = \frac{1}{2}m_1 v_{1f}^2 = \frac{1}{2}m_1 \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 v_{1i}^2 \] The fractional kinetic energy lost is: \[ f_1 = \frac{K_{1f}}{K_{1i}} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 \] The fractional kinetic energy gained by the target nucleus: \[ f_2 = 1 - f_1 = \frac{4m_1 m_2}{(m_1 + m_2)^2} \]
For deuterium (\(m_2 = 2m_1\)): \[ f_1 = \left(\frac{m_1 - 2m_1}{m_1 + 2m_1}\right)^2 = \left(\frac{-1}{3}\right)^2 = \frac{1}{9} \approx 11\% \] So \(f_2 = 1 - \frac{1}{9} = \frac{8}{9} \approx 89\%\) of the neutron's energy is transferred to deuterium.

For carbon (\(f_1 = 71.6\%\) and \(f_2 = 28.4\%\)), in practice \(f_1\) is small and head-on collisions are rare.